Question

In a compound microscope, the focal length of the objective lens and the eye lens are $4\,mm$ and $25\,mm$ and the length of the tube is $16\,cm$ . For relaxed eye position, find its magnifying power.

• A. $32.75$
• B. $327.5$
• C. $0.3275$
• D. None of the above

Answer 1

Answer: (B)

$f_{o}=4\,mm=0.4\,cm$
$f_{e}=25\,mm=2.5\,cm$
Length of the tube, $L_{t}$ = $16\,cm$
Magnifying power for relaxed eye position is given by:
$m=\frac{\left(L_{t} - f_{o} - f_{e}\right) D}{f_{o} f_{e}}$
$m=\frac{\left(\right. 16 - 0 . 4 - 2 . 5 \left.\right) \times 25}{0 . 4 \times 2 . 5}$
$m=327.5$